Determine all the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for every $x \in \mathbb{R}$, $f(2x) = 2f(x)$ . Basically I thought about a kind of modulo 2 equivalence class for real numbers, if that makes sense. With that, and noting that for each number $y \in [1,2)$, the numbers $2y$ and $y/2 $ are not in $ [1,2)$, I have defined an arbitrary function on the interval $[1,2)$ and repeated it (stretched or shrunk in a way that maintains the property) along the real axis. I have found that, if $h$ and $g$ are arbitrary functions from $\mathbb{R}$ to $\mathbb{R}$, then an $f$ constructed as follows satisfies the request. $\lfloor \cdot \rfloor$ represents the floor function.
\begin{equation*}
  f(x) = \begin{cases}
  2^{\lfloor\log_2 |x|\rfloor} \cdot  h(|x| \cdot 2^{-\lfloor\log_2 |x|-1\rfloor} )& x >0 \\
  2^{\lfloor\log_2 |x|\rfloor} \cdot  g(|x| \cdot 2^{-\lfloor\log_2 |x|-1\rfloor} ) & x <0\\
  0    & x =0 
  \end{cases}
 \end{equation*}
This works because if $x>0$, then: $$f(2x) = 2^{\lfloor\log_2 (2x)\rfloor} \cdot  h(2x \cdot 2^{-\lfloor\log_2 (2x)-1\rfloor} ) = 2^{1 + \lfloor\log_2 (x)\rfloor} \cdot  h(2x \cdot 2^{- 1 -\lfloor\log_2 (x)-1\rfloor} ) = 2f(x) \, .$$
Similarly, this happens when $x<0$, and by defining $f(0)=0$ we also get that the property is satisfied for $x=0$.
My question is this: how can it be said that these are all the possible functions that satisfy the request? And also, if $h$ and $g$ are not the same linear function, can $f$ be differentiable at $x=0$? It seems strange to me that a function that oscillates infinitely many times near the origin would be differentiable there. Can it be that, if neither $g$ nor $h$ are linear, $f$ is $C^\infty$ at every $x\neq 0$? My candidate for that is $g(x)=h(x)=\exp\left[-\left(\frac{1}{|x-2|} + \frac{1}{|x-4|} \right)\right]$ . Lastly, is there a prettier way to express $f$ ?
Thanks for reading :)
 A: It's easy to prove that what you wrote is the only correct form (and I don't think it can get much prettier). In fact every f such that $f(2x)=2f(x)$ of course induces an $h:[1,2) \to \mathbb{R}$ by restriction; then for every $x >0$ there exists a unique $k \in \mathbb{Z} $ such that $2^k x \in [1,2)$ and in particular $f(x)=2^{-k} f( 2^k x) = 2^{-k}h(2^kx)$ as you described.
regarding the differentiability in $0$ one has, for every $x \in [1,2)$ that $x_k=x \cdot 2^{-k}$ is a sequence tending to $0$. But then, by the properties of $f$, if it is differentiable in $0$ we would have
$$f'(0) = \lim_{k \to \infty} \frac {f(x_k)-f(0)}{x_k-0} = \frac {h(x)}{x}$$
and the same should be true for $g$; in particular $f$ is differentiable in $0$ iff $h(x)=ax$ and $g(x)=ax$ for the same $a \in \mathbb{R}$, which will be the derivative of $f$ in $0$.
You can have that the function is $C^{\infty}$ outside $0$, it is sufficient to take any $C^{\infty}$ function $t$ which is $0$ outside $[0,1]$, for example $t(x)=e^{- 1/|x(1-x)|}$, similarly as you suggested...
then you can consider $h(x)=g(x)=x+ t(2*(x-4/3))$ (in such a way that it gets modified only on $[4/3,11/6]$, and there should be no problems at $1$ and $2$, but you can take also $h(x)=x+t(x-1)$ it works fine).
In conclusion, in fact everything you said is true!
